Bull. Korean Math. Soc. 50 (2013), No. 2, pp. 601–610 http://dx.doi.org/10.4134/BKMS.2013.50.2.601
A CLASS OF ARITHMETIC FUNCTIONS ON PSL2(Z)
Paul Spiegelhalter and Alexandru Zaharescu
Abstract. In [3] and [2], Atanassov introduced the two arithmetic func- tions
I(n) = Y
pα||n
p1/α and R(n) = Y
pα||n
pα−1
called the irrational factor and the restrictive factor, respectively. Alkan, Ledoan, Panaitopol, and the authors explore properties of these arith- metic functions in [1], [7], [8] and [9]. In the present paper, we generalize these functions to a larger class of elements of PSL2(Z), and explore some of the properties of these maps.
1. Introduction
In [3] and [2], Atanassov introduced the two arithmetic functions I(n) = Y
pα||n
p1/α and R(n) = Y
pα||n
pα−1
called the irrational factor and the strong restrictive factor, respectively. These functions are multiplicative, and satisfy the inequality
I(n)R(n)2≥ n,
with equality if and only if n is squarefree. In [8], Panaitopol showed that X∞
n=1
1
I(n)R(n)φ(n) < e2, and proved that the function
G(n) = Yn ν=1
I(ν)1/n satisfies the inequalities
e−7n < G(n) < n.
Received December 4, 2011.
2010 Mathematics Subject Classification. Primary 11N37; Secondary 11B57.
Key words and phrases. PSL2(Z), Farey fractions, Dirichlet series.
The second author is supported in part by NSF grant DMS-0901621.
2013 The Korean Mathematical Societyc 601
In [1], Alkan, Ledoan and one of the authors describe a precise asymptotic for the function G(n), and establish further results showing that the function I(n) is very regular on average.
In [7], asymptotic formulas are established for certain weighted real moments of the restrictive factor R(n). In [9], the authors establish asymptotic formulas for weighted combinations I(n)αR(n)β.
In the present paper we consider for a matrix A =
a b c d
in PSL2(Z) the fractional linear transformation Az given by Az = az + b
cz + d. For each positive integer n, define
fA(n) = Y
pα||n
paα+bcα+d.
As an example, the function I(n) is equal to fA0(n) for A0=
0 1 1 0
.
We shall consider weighted averages of the functions fA(n). Let MA(x) = 1
x X
n≤x
1 −n x
fA(n).
Consider the subset A ⊂ PSL2(Z) given by A =
A =
a b c d
∈ PSL2(Z) : det A = −1, a, b, d ≥ 0, c ≥ 1
. Define for each positive rational number r
Er= {A ∈ A : MA(x) ≍ xr as x → ∞}.
Note that if r1 6= r2, then Er1 ∩ Er2 = Ø. We will prove that each Er with r > 0 consists of exactly one element.
For each matrix A in A we define the associated series (An)n∈Nby An = An =an + b
cn + d.
As we shall see, the associated series plays an important role in our computa- tions. Clearly, if A ∈ A, then An is monotone decreasing and has the finite limit A∞:= a/c.
We have the following result.
Theorem 1.1. Given A ∈ A, if A1 > 0, then there are positive real-valued constants KA andc such that
MA(x) = KAxA1+ OA
xA1−1/2exp{−c(log x)3/5(log log x)−1/5} . We remark that under the Riemann hypothesis, for a restricted class of matrices one has an asymptotic formula for the error term in Theorem 1.1 of the form
(1) MA(x) − KAxA1∼ eKAx12(A2−1)
for a real-valued constant eKA. This naturally leads one to consider the maps ψj : A → Q+ for j = 1, 2 given by
(2) ψj(A) = Aj.
Since, as mentioned above, each Er consists of exactly one element, it follows that there is a well-defined map s : Q+→ Q+given by
(3) s(r) = ψ2◦ ψ−11 (r).
The map s(r) tells us how accurately the main term KAxA1 approximates MA(x) in (1), in the sense that it gives the exact order of magnitude of the error MA(x) − KAxA1.
Although it can be shown that this map is nowhere continuous, one can obtain asymptotic formulas for the average value of s(r), with r in various ranges. For example, define the height function for each rational r = p/q with q ≥ 1 and (p, q) = 1 by
h(r) := max{|p|, |q|}.
We have the following result.
Theorem 1.2. For any δ > 0, X
r∈Q+∩[0,1]
h(r)≤X
s(r) = 3
2π2X2+ Oδ(X11/6+δ).
2. Asymptotics of the average Consider the Dirichlet series
FA(s) = X∞ n=1
fA(n) ns .
We will take advantage of the meromorphic continuation of FA(s) in the case where det A = −1.
Proof of Theorem 1.1. We prove the result with
KA= 1
(1 + A1)(2 + A1)ζ(2)TA(1 + A1).
If det A = −1, then pAα≤ pA1 for all α ≥ 1, so fA(n) ≤ nA1, hence FA(s) converges in the half plane ℜs = σ > 1 + A1. Moreover, FA(s) has an Euler product in that region. Write
FA(s) = ζ(s − A1) ζ(2s − 2A1)
Y
p
(1 + gp(s)) ,
where
gp(s) =
1 + pA1
ps
−1 ∞X
k=2
pAk pks.
Note that if det(A) = −1, then A1− A2 = (c+d)(2c+d)1 ≤ 12. Take ǫ > 0. For σ ≥ A1+ ǫ we have
1 + pA1
ps
−1
≪ǫ1.
Also, for σ ≥ 12(1 + A2+ ǫ) we have X∞ k=2
pAk pks ≪ pA2
p2s ≪ǫ 1 p1+ǫ. Thus for σ ≥ max{A1+ ǫ,12(1 + A2+ ǫ)} the sumP
p|gp(s)| converges, hence TA(s) =Y
p
(1 + gp(s))
is analytic for σ > σ0= max
A1,12(1 + A2)
, so FA(s) is meromorphic there, with a poles at s = 1 + A1.
To continue, we utilize a variant of Perron’s formula and write X
n≤x
1 − n x
fA(n) = 1 2πi
Z c+i∞
c−i∞
ζ(s − A1)
ζ(2s − 2A1)TA(s) xs s(s + 1)ds, where 1 + A1< c ≤ 5/4 + A1.
We apply the zero-free region for ζ(s) due to Korobov [6] and Vinogradov [12] (see Chapters 2 and 5 of the reference by Walfisz [13] for an alternative treatment)
σ ≥ 1 − c0(log t)−2/3(log log t)−1/3 for t ≥ t0, in which
1
|ζ(s)| ≪ (log t)2/3(log log t)1/3. Fix 0 < U < T ≤ x, let ν = 1/2 + A1 and
η = ν − c0(log U )−2/3(log log U )−1/3.
Deform the path of integration into the union of the line segments
γ1, γ9: s = c + it if |t| ≥ T γ2, γ8: s = σ ± iT if ν ≤ σ ≤ c γ3, γ7: s = ν + it if U ≤ |t| ≤ T γ4, γ6: s = σ ± iU if η ≤ σ ≤ ν γ5: s = η + it if |t| ≤ U.
The integrand is analytic on and within this modified contour, hence by Cauchy’s theorem
xMA(x) = 1
(1 + A1)(2 + A1)ζ(2)TA(1 + A1)x1+A1+ X9 k=1
Jk,
with the main terms coming from the residue at the simple pole at s = 1 + A1. In order to estimate the integral along our modified contour we will make use of the bounds
|ζ(σ + it)| =
O(t(1−σ)/2, if 0 ≤ σ ≤ 1 and |t| ≥ 1 O(log t), if 1 ≤ σ ≤ 2
O(1), if σ ≥ 2 (see [11], §3.11 and §5.1).
On the line segments on which s = c + it, |t| ≥ T , we have that ζ(s − A1) ≪ log t and 1/ζ(2s − 2A1) ≪ log t, so
|J1|, |J9| ≪ Z ∞
T
(log t)2 xc
|(c + it)(c + 1 + it)|dt
≪ xc(log T )2
T .
On the line segments on which s = σ + iT , ν ≤ σ ≤ c, we have that 1/ζ(2s − 2A1) ≪ log T , ζ(s − A1) ≪ T(1−σ+A1)/2 for ν ≤ σ ≤ 1 + A1, and ζ(s − A1) ≪ log T for 1 + A1≤ σ ≤ c. So
|J2|, |J8| ≪
Z 1+A1
ν
T12(1−σ+A1)log Txσ T2dσ +
Z c 1+A1
(log T )2xσ T2dσ
≪ T12(1+A1)log T maxn x
√T
ν
,
x
√T
1+A1o
+(log T )2 T2 xc. On the line segments on which s = ν + it, U ≤ |t| ≤ T , we have that ζ(s − A1) ≪ t(1−ν+A1)/2 and 1/ζ(2s − 2A1) ≪ log t, so
|J3|, |J7| ≪ Z T
U
(log t)t12(1−ν+A1) xν
|(ν + it)(ν + 1 + it)|dt
≪ log T U3/4xν.
On the line segments on which s = σ +iU , η ≤ σ ≤ ν, we have that ζ(s−A1) ≪ U(1−σ+A1)/2 and 1/ζ(2s − 2A1) ≪ log U, so
|J4|, |J6| ≪ Z ν
η
(log U )U12(1−σ+A1)xσ U2 dσ
≪ U12(1+A1)−2log U maxn x
√U
ν
,
x
√U
ηo .
On the line segment on which s = η + it, |t| ≤ U, we have that ζ(s − A1) ≪ (|t| + 1)(1−η+A1)/2and 1/ζ(2s − 2A1) ≪ log(|t| + 1), so
|J5| ≪ Z U
−U(|t| + 1)1−η+A1log(|t| + 1) xη
|η + it||η + 1 + it| dt
≪ xη Z U
−U(|t| + 1)12(1−η+A1)−2log(|t| + 1) dt.
Since 12(1 − η + A1) − 2 ≤ −32 for U sufficiently large, the above integral converges, hence |J5| ≪ xη.
We collect all estimates, and take T = x2 and
U = exp{c2(log x)3/5(log log x)−1/5}
to obtain the desired result.
One could instead factor
1 + pA1
ps +pA2 p2s +pA3
p3s + · · ·
=
1 +pA1
ps
1 +pA2
p2s
(1 + gp(s)) with
gp(s) =
1 + pA1
ps
−1 1 + pA2
p2s
−1
−pA1+A2 p3s +
X∞ k=3
pAk pks
!
so that
FA(s) = ζ(s − A1) ζ(2s − 2A1)
ζ(2s − A2) ζ(4s − 2A2)
Y
p
(1 + gp(s)) .
Under the Riemann hypothesis, we get a second order term of the form eKAxA2 in the asymptotic formula for FA(s) provided that 14+ A1< 12(1 + A2). That is, provided that
a + b < c + d
2 − 1
2c + d.
This occurs for matrices A in A with restrictions on c and d. One can see that A1will lie in the interval (0, 1/2).
3. Mapping through PSL2(Z) We now return to the two maps ψ1 and ψ2defined in (2).
Lemma 3.1. The map ψ1 is bijective.
Proof. For pq ∈ Q+, consider the set of matrixes A =
a b c d
in A such that ψ1(A) = pq. We note that any such quadruple (a, b, c, d) is constrained by c ≥ 0, d ≥ 0,
(4) ad − bc = −1
and
(5) c + d = q
p(a + b)
(Note that (4) implies that p cannot be zero). By (5) we have c = q
p(a + b) − d.
Inserting this into (4) gives us
ad − b(a + b)q
p+ bd = −1 so
(a + b)(pd − qb) = −p.
Write a + b = ±n for some positive integer n | p. By (5) we have c + d =
q
p(±n) ∈ Z so p | n, hence p = n.
There are two cases: If a + b = −p, then c + d = −q. This contradicts the assumptions that q ≥ 1 and c and d are non-negative. On the other hand, if a + b = p, then c + d = q, so (4) gives us
a(q − c) − bc = −1 so
(6) pc = 1 + aq.
So c is uniquely determined by cp ≡ 1 (mod q) and 1 ≤ c < q. Then d is uniquely determined by d = q − c, and a and b by a = 1−pcq and b = p − a. In the case where p/q ∈ (0, 1], we identify p/q as an element of FQ, the Farey fractions of order Q, with Q ≥ q. If we consider the “minimal” set of Farey fractions Fq containing p/q, then elementary properties of Farey fractions (see for example Chapter 3 of [5]) give that the adjacent Farey fractions p′/q′ <
p/q < p′′/q′′ satisfy q′ = ¯p, p′ = ¯q, p′′= p − ¯q and q′′ = q − ¯p. Here ¯p is the
unique integer 1 ≤ ¯p < q satisfying p¯p ≡ 1 (mod q) and ¯q is the unique integer 1 ≤ ¯q < p satisfying q¯q ≡ 1 (mod p). We can write
ψ1(p/q) =
q¯ p − ¯q
¯ p q − ¯p
.
That is, the matrix ψ1(p/q) is comprised of the “parent” Farey fractions in Fq−1.
Additionally, we can write the function s(p/q) from (3) uniquely as
(7) s(p/q) = pp − 1 + pq¯
q(¯p + q) .
To prove Theorem 1.2, we will use the following result (see Lemma 2.3 of [4]).
Lemma 3.2. Assume that q ≥ 1 and h are two given integers, I and J are intervals of length less than q, and f : I × J → R is a C1 function. Then for any integer T ≥ 1 and any δ > 0
X
a∈I,b∈J ab≡h (mod q)
gcd(a,b)=1
f (a, b) = φ(q) q2
ZZ
I×Jf (x, y)dxdy + E,
with
E ≪δ T2kfk∞q1/2+δgcd(h, q)1/2
+ T k∇fk∞q3/2+δgcd(h, q)1/2+k∇fk∞|I||J |
T ,
wherekfk∞ andk∇fk∞denote the sup-norm off and respectively |∂f∂x| + |∂f∂y| on I × J .
Proof of Theorem 1.2. Let Q = ⌊X⌋. Since r ∈ FQ we have X
r∈Q+∩[0,1]
h(r)≤X
s(r) = X
1≤q≤Q 1≤p<q (p,q)=1
s(p/q).
We use (7) and Lemma 3.2 with T = q16−δ3 to get that the right-hand sum is equal to
X
1≤q≤Q
X
1≤p<q 1≤ ¯p<q p ¯p≡1 (mod q)
(p,q)=1
pp − 1 + pq¯
q(¯p + q) = X
1≤q≤Q
φ(q) q2
Z Z
[1,q)2
vu − 1 + uq
q(v + q) dudv + E
= X
1≤q≤Q
φ(q) Z Z
[1/q,1]2
xy − q12 + x
y + 1 dxdy + E.
where E ≪δ q5/6+δ. The integral is equal to 1
2
1 − 1
q2
1 −1
q
−q − 1 q3
log 2 − log
1 + 1
q
= 1 2+ O
1 q
so X
r∈Q+∩[0,1]
h(r)≤X
s(r) = 1 2
X
1≤q≤Q
φ(q) + O
X
1≤q≤Q
φ(q) q
+ O
X
1≤q≤Q
q5/6+δ
.
One can use the methods of Section 2 to estimate the sums over φ(q), or use partial summation along with standard estimates (see for example [13] or Chapter 18 of [5]). This gives the main term of our theorem; the first error term above is O(X), and the second is Oδ(X11/6+δ).
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Paul Spiegelhalter
Department of Mathematics University of Illinois
1409 West Green Street, Urbana, IL 61801, USA E-mail address: [email protected]
Alexandru Zaharescu Department of Mathematics University of Illinois
1409 West Green Street, Urbana, IL 61801, USA E-mail address: [email protected]